Notes · MathematicsUK · A-Levels

Statistical hypothesis testing

Hypothesis testing decides, from sample evidence, between two competing claims about a population. The A-Level covers the logic of testing (null and alternative hypotheses, significance levels, one- and two-tailed tests) and three specific tests: for a binomial proportion, for a correlation coefficient, and for the mean of a normal distribution with known variance.

4 sections~11 min reading time3 competenciesLevel Standard 1 · Advanced 3

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Exam profile

AO1 · Carry out hypothesis tests and find critical regionsAO2 · Interpret and communicate conclusions in contextAO3 · Formulate hypotheses appropriate to a stated question

Operators:testfindstateconcludedetermineshow that

basic level

AS-Level covers the language of hypothesis testing and the test for a binomial proportion.

higher level

The full A-Level adds tests for a correlation coefficient and for the mean of a normal distribution with known variance.

Depth

Text

Contents · 4 sections

Statistical hypothesis testing

The logic of hypothesis testing#

StandardDfE GCE Mathematics — testingAQA 7357 (Statistics)

Two-tailed critical regions

Fig. 1For a two-tailed test the significance level is split between the two tails; a test statistic in either tail rejects

H_0

A hypothesis test weighs sample evidence against a default claim. The null hypothesis

H_0

states the status quo (a specific value of a parameter, such as

p = 0.5

), and the alternative hypothesis

H_1

states what you suspect instead. The test asks: if

H_0

were true, how likely is the observed data? If the data would be very unlikely under

H_0

, we reject

H_0

in favour of

H_1

; otherwise we do not.

'Very unlikely' is made precise by the significance level, usually 5% or 1%. This is the probability, chosen in advance, of rejecting

H_0

when it is in fact true (a Type I error). A smaller significance level demands stronger evidence to reject

H_0

. The corresponding critical region is the set of outcomes so extreme that observing one leads to rejection.

The direction of the alternative determines whether the test is one- or two-tailed. If

H_1

specifies a direction (

p > 0.5

p < 0.5

) the test is one-tailed and the critical region sits in one tail; if

H_1

is 'not equal' (

p \neq 0.5

) the test is two-tailed and the significance level is split between both tails. Choosing the correct type from the wording of the question is essential and frequently examined.

There are two equivalent methods. The critical-region method finds the extreme outcomes in advance and checks whether the data fall in them; the

p

-value method computes the probability of a result at least as extreme as observed and compares it with the significance level. Either way, the conclusion must be stated in context and never as certainty — a test provides evidence, not proof.

\text{reject } H_0 \iff p\text{-value} < \text{significance level}

The decision rule

If a result this extreme is too unlikely under $H_0$ , reject it; otherwise there is insufficient evidence.

Worked example

Setting up a test

A coin is suspected of being biased towards heads. It is tossed 20 times. Write suitable hypotheses and state the type of test.

Define the parameter
Let $p$ be the probability of a head.
State the hypotheses
$H_0: p = 0.5$ (fair) against $H_1: p > 0.5$ (biased towards heads).
Type of test
The alternative gives a direction, so the test is one-tailed (upper tail).

Result: $H_0: p = 0.5$ , $H_1: p > 0.5$ ; a one-tailed test.

Typical mistakes

Using a one-tailed test when the alternative is 'not equal' (which needs two tails), or vice versa.
Stating the conclusion as certainty ('the coin is fair') rather than as evidence.

Active revision

A manufacturer claims 20% of chocolates are dark. A researcher suspects the proportion is different. State suitable hypotheses and say whether the test is one- or two-tailed.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Testing a binomial proportion#

AdvancedDfE GCE Mathematics — testingAQA 7357 (Statistics)

Binomial distribution under the null hypothesis

Fig. 2Under

H_0: p = 0.5

with

n = 10

, the upper-tail values (8, 9, 10) make up the critical region for an upper-tailed test.

A binomial test decides whether a population proportion

p

differs from a claimed value. Under

H_0

the number of successes

X

n

trials follows

B(n, p_0)

, and this known distribution provides the yardstick against which the observed count is judged. The test statistic is simply the observed number of successes.

In the

p

-value method you compute the probability, under

H_0

, of a result at least as extreme as the one observed, and compare it with the significance level. For an upper-tail test with observed value

x

you find

P(X \geq x)

; for a lower-tail test,

P(X \leq x)

; for a two-tailed test you work in the relevant tail and compare with half the significance level. If this probability is smaller than the level, the result is significant and

H_0

is rejected.

In the critical-region method you find, before looking at the data, the set of extreme values whose total probability is at most the significance level. Because

X

is discrete, the critical region rarely has probability exactly equal to the level; you take the largest region not exceeding it, and its actual probability is the true significance (or 'actual significance level') of the test.

Whatever the method, the conclusion is a statement in context: 'there is sufficient evidence at the 5% level to reject

H_0

and conclude the proportion has increased', or 'insufficient evidence to reject

H_0

'. Naming the significance level and referring to the original claim are required for full marks.

\text{Under } H_0,\ X \sim B(n, p_0); \quad \text{reject if } P(X \geq x) < \alpha

Upper-tail binomial test

Compare the tail probability of the observed count with the significance level $\alpha$ .

Worked example

A one-tailed binomial test

A treatment is claimed to succeed 40% of the time. In a trial of 20 patients, 13 recover. Test at the 5% level whether the success rate exceeds 40%.

Hypotheses
$H_0: p = 0.4$ , $H_1: p > 0.4$ ; let $X \sim B(20, 0.4)$ under $H_0$ .
Tail probability
$P(X \geq 13) = 1 - P(X \leq 12) = 1 - 0.9790 = 0.0210$ .
$P(X \geq 13) = 1 - 0.9790 = 0.0210$
Compare and conclude
$0.0210 < 0.05$ , so the result is significant.

Result: Since $0.0210 < 0.05$ , reject $H_0$ : there is evidence at the 5% level that the success rate exceeds 40%.

Typical mistakes

Computing $P(X > x)$ instead of $P(X \geq x)$ (missing the observed value itself).
For a two-tailed test, comparing the tail probability with the full level instead of half.

Active revision

It is claimed that 30% of customers choose the vegetarian option. In a sample of 25, 12 do. Test at the 5% level whether the proportion is greater than 30%.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Correlation hypothesis tests#

AdvancedDfE GCE Mathematics — testingAQA 7357 (Statistics)

Comparing r with its critical value

Fig. 3If the sample

r

exceeds the critical value (here

0.55

) the result is significant and

H_0: \rho = 0

is rejected.

A correlation test decides whether a non-zero correlation observed in a sample is strong enough to be evidence of correlation in the population, or could plausibly have arisen by chance from an uncorrelated population. The population correlation coefficient is denoted

\rho

, and the sample gives the product moment correlation coefficient

r

. The null hypothesis is always

H_0: \rho = 0

(no correlation).

The alternative reflects what is being tested:

H_1: \rho > 0

(positive correlation),

H_1: \rho < 0

(negative correlation), or

H_1: \rho \neq 0

(some correlation), giving a one- or two-tailed test. The choice, as always, comes from the wording of the question and determines how the significance level is applied.

The test compares the sample

r

with a critical value obtained from tables for the given sample size

n

and significance level. If the sample correlation is more extreme than the critical value —

|r|

larger than the tabulated figure — the result is significant and

H_0

is rejected. Larger samples require a smaller

r

to be significant, because a given correlation is less likely to arise by chance from more data.

The conclusion is stated in context: 'there is evidence at the 5% level of positive correlation between rainfall and yield'. As with all correlation, significance does not establish causation, and a careful answer says only that the data provide evidence of an association, not that one variable causes the other.

H_0: \rho = 0; \quad \text{reject if } |r| > r_{\text{critical}}(n, \alpha)

The correlation test

Compare the sample $r$ with the tabulated critical value for the sample size and significance level.

Worked example

A correlation significance test

A sample of 10 pairs gives $r = 0.63$ . Test at the 5% level whether there is positive correlation, given the one-tailed critical value is $0.5494$ .

Hypotheses
$H_0: \rho = 0$ , $H_1: \rho > 0$ (one-tailed).
Compare
$r = 0.63 > 0.5494$ , the critical value.
$0.63 > 0.5494$
Conclude
The sample correlation exceeds the critical value, so the result is significant.

Result: Since $0.63 > 0.5494$ , reject $H_0$ : there is evidence at the 5% level of positive correlation.

Typical mistakes

Writing the hypotheses in terms of $r$ rather than the population parameter $\rho$ .
Using the wrong critical value for the sample size or number of tails.

Active revision

For a sample of $n = 15$ pairs, the correlation is $r = 0.48$ . The critical value for a one-tailed 5% test is $0.4409$ . Test at the 5% level for positive correlation.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Testing a normal mean#

AdvancedDfE GCE Mathematics — testingAQA 7357 (Statistics)

Rejection regions for a two-tailed z-test

Fig. 4For a two-tailed 5% test, reject

H_0

if the standardised statistic lies beyond

\pm 1.96

(shaded tails).

A test for the mean of a normal distribution decides whether the population mean

\mu

differs from a claimed value

\mu_0

, based on a sample mean

\bar{x}

. The key theoretical result is that if individual values come from

N(\mu, \sigma^2)

, then the mean of a sample of size

n

is itself normally distributed:

\bar{X} \sim N\!\left(\mu, \dfrac{\sigma^2}{n}\right)

. The spread of the sample mean is smaller than that of individuals by a factor of

\sqrt{n}

Under

H_0: \mu = \mu_0

, the standardised test statistic is

Z = \dfrac{\bar{X} - \mu_0}{\sigma / \sqrt{n}}

, which follows the standard normal

N(0, 1)

. The observed sample mean is converted to a

z

-value, and either its tail probability (

p

-value) is compared with the significance level, or the

z

-value is compared with the critical value (

\pm 1.96

for a two-tailed 5% test,

1.645

for one tail).

The direction of the alternative again fixes the tail(s):

H_1: \mu > \mu_0

\mu < \mu_0

\mu \neq \mu_0

. For this test the population variance

\sigma^2

is assumed known; the sample provides only the mean

\bar{x}

, and the known standard deviation of the sample mean,

\dfrac{\sigma}{\sqrt{n}}

, is the yardstick.

As with every test, the conclusion is a contextual statement about the evidence: 'there is sufficient evidence at the 5% level that the mean fill weight has fallen below 500 g'. Sketching the standard normal curve, shading the rejection region and marking the test statistic makes the comparison unambiguous and guards against tail errors.

\bar{X} \sim N\!\left(\mu, \frac{\sigma^2}{n}\right), \qquad Z = \frac{\bar{X} - \mu_0}{\sigma / \sqrt{n}} \sim N(0, 1)

Distribution of the sample mean and the test statistic

The sample mean is normal with reduced variance; standardising gives the z test statistic.

Worked example

A test for a normal mean

A machine fills bottles to $N(500, 4^2)$ ml. A sample of 25 bottles has mean 498.5 ml. Test at the 5% level whether the mean has decreased.

Hypotheses
$H_0: \mu = 500$ , $H_1: \mu < 500$ (one-tailed); $\bar{X} \sim N\!\left(500, \tfrac{16}{25}\right)$ .
Standardise
$Z = \dfrac{498.5 - 500}{4/\sqrt{25}} = \dfrac{-1.5}{0.8} = -1.875$ .
$Z = \frac{498.5 - 500}{4/5} = -1.875$
Compare with the critical value
The one-tailed 5% critical value is $-1.645$ ; since $-1.875 < -1.645$ , the statistic is in the rejection region.

Result: Since $-1.875 < -1.645$ , reject $H_0$ : there is evidence at the 5% level that the mean has decreased.

Typical mistakes

Dividing by $\sigma$ instead of $\dfrac{\sigma}{\sqrt{n}}$ when standardising the sample mean.
Using the two-tailed critical value for a one-tailed test.

Active revision

The mean mass of a bag of sugar should be 1000 g with standard deviation 8 g. A sample of 16 bags has mean 996 g. Test at the 5% level whether the mean has fallen below 1000 g.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Statistical hypothesis testing

4 sections~11 min reading time3 competenciesLevel Standard 1 · Advanced 3

Statistical hypothesis testing

Key points

Setting up a test

Key points

A one-tailed binomial test

Key points

A correlation significance test

Key points

A test for a normal mean

Sources

Statistical hypothesis testing

Key points

Setting up a test

Key points

A one-tailed binomial test

Key points

A correlation significance test

Key points

A test for a normal mean

Sources